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I'd like to have a clear idea on how to solve this (seems to be very simple, but I haven't seen something like that yet).

Consider $\mathbb{R}^{2}$ with the metric given by $ds²$ = $\frac{4(dx² + dy²)}{1+x²+y²}$. Compute the geodesic curvatures of the following curves:

a) $y = x$

b) $y = x+1$

c) $x² + y² = 1$

Or maybe if there is a general idea for this kind of problem. Thanks in advance :)

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Where did you find this problem? Also, what is your definition of "geodesic curvature"? (There are many equivalent definitions out there.) Do you have a formula for it? – Jesse Madnick Jan 30 '13 at 4:25
up vote 1 down vote accepted

Given a bit more background (i.e. what level you are out and how this problem arised), the following answer could be altered. It appears that this is probably an undergraduate or beginning graduate level course on differential geometry, and I will proceed accordingly. The approach taken below is a computational one and in the end, it will note assume that your curves are parametrized by arc length.

From a straight computational standpoint, the metric given is very nice; Relative to the semi-standard notation $Edx^2 + 2Fdxdy + Gdy^2$, we have $E = G$ and $F = 0$ and the corresponding Christoffel Symbols of the second are very nice:

\begin{align*} \Gamma^{1}_{\,11} &= \frac{E_x}{2E} \quad\quad\quad &\Gamma^{2}_{\,11} = -\frac{E_y}{2G}\\ \Gamma^{1}_{\,12} &= \frac{E_y}{2E} \quad\quad\quad &\Gamma^{2}_{\,12} = \frac{G_x}{2G}\\ \Gamma^{1}_{\,22} &= -\frac{G_x}{2E} \quad\quad\quad &\Gamma^{2}_{\,22} = \frac{G_y}{2G},\\ \end{align*} with the usual symmetries $\Gamma^{i}_{\,12} = \Gamma^{i}_{\,21}$, $i = 1..2$.

Computing the Christoffel Symbols will give you your connection (or covariant derivative) $\nabla$.

The geodesic curvature of a curve $c : [a, b] \to M = \mathbb{R}^{2}_{g=ds^2}$ at a point c(t), where c is parametrized by arc length, is length of the vector $\frac{Dc^\prime}{dt}(t)$ with respect to the given metric (where $\frac{D}{dt}$ is the covariant differentiation along the curve). Given the fact that the tangent vector fields along the curves in question are all extendable to a neighborhood of the curve, then you can use the formula $$\frac{Dc^\prime}{dt} = \nabla_{c^{\prime}(t)}c^\prime(t),$$ and expand out the right hand side with the help of the $\Gamma^{i}_{\,jk}.$

With all of this being said, however, the above assumes that the curve $c$ is parametrized by arc length. I have no idea how easy (or difficult) it is to parametrize the indicated curves by arc length with respect to the given metric $ds^2$. Since this is the special case of surface, there does appear to be a special formula for the case when the curve is not parametrized by arc length. See the following.

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